A Note on the Arbelos in Wasan Geometry, Satoh's Problem

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Sangaku Journal of Mathematics (SJM) ⃝c SJM ISSN 2534-9562 Volume 3 (2019), pp.15-16 Received 13 February 2019. Published on-line 15 February 2019 web: http://www.sangaku-journal.eu/ ⃝c The Author(s) This article is published with open access1.

A note on the arbelos in Wasan geometry, Satoh’s problem

Hiroshi Okumura Maebashi Gunma 371-0123, Japan e-mail: [email protected]

Abstract. We generalize a sangaku problem involving an arbelos proposed by Satoh, and show the existence of three congruent non-Archimedean circles. Keywords. arbelos, non-Archimedean congruent circles. Mathematics Subject Classiﬁcation (2010). 01A27, 51M04

1. Introduction

We consider the arbelos appeared in Wasan geometry, and consider an arbelos formed by three semicircles α, β and γ with diameters AO, BO and AB, respectively for a point O on the segment AB (see Figure 1). For a point F on the segment AO, let h be the perpendicular to AB at F . The circle with diameter FO is denoted by δ1, and the incircle of the curvilinear triangle made by α, γ and h is denoted by δ2. Let ri be the radius of δi. The incircle of the curvilinear triangle made by α, γ and the external common tangent of α and β is denoted by ε. Let e be the radius of ε. In this note we generalize the following problem in the sangaku hung in Fukushima in 1883 proposed by Satoh (佐藤金之明)[2] (see Figure 2).

γ γ δ 2 δ2 ε α α ε β β h h δ1 δ1 B O F A B O F A Figure 1. Figure 2.

Problem 1. If α and β are congruent and δ1 and δ2 are also congruent, show r1 = 3e. 1This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited. 15 16 A note on the arbelos in Wasan geometry, Satoh’s problem

2. Generalization

We give a condition that the circles δ1 and δ2 are congruent, and show the existence of two pairs of three congruent circles. The radii of α and β are denoted by a and b, respectively. Let the tangent of α from B touch α at a point T , and let U be the foot of perpendicular from T to AB and u = ab/(a + 2b) (see Figure 3). Proposition 1. |OU| = 2u. Proof. If C is the center of α, the triangles BCT and TCU are similar. Hence we get (a − |OU|)/a = a/(a + 2b), which implies |OU| = 2u. □

Theorem 1. The circles δ1 and δ2 are congruent if and only if h passes through T . In this event the common radius equals u.

Proof. Let k = |FO|. Since r2 is proportional to the distance between the center of δ2 and the radical axis of α and γ [1, p. 108], r2/(2a − k) = b/(2a + 2b), i.e., r2 = (2a−k)b/(2a+2b), while r1 = k/2. Therefore r1 = r2 is equivalent to k = 2u (see Figure 4). □

γ γ

δ2

T α T α h β β

δ1 B O U C A B O F =U A Figure 3. Figure 4.

Since e = a2b/(a + 2b)2 [3, 4], we get u = e(a + 2b)/a. This gives a solution of Problem 1 in the case a ≠ b and implies u = 3e in the case a = b. The circle touching the tangent of β from A and h at F from the side opposite to B is congruent to δ2 [3]. Circles of radius ab/(a + b) are said to be Archimedean. Therefore we get three congruent non-Archimedean circles of radius u in the case F = U. Exchanging the roles of α and β we get another three congruent circles of radius ab/(2a + b), which are denoted by the green circles in Figure 4.

References

[1] J. L. Coolidge, A treatise on the circle and the sphere, Chelsea, New York, 1971 (reprint of 1916 edition). [2] Fukushimaken Wasan Kenky¯uHozonkai (福島県和算研究保存会) ed., The sangaku in Fukushima (福島の算額), S¯ojuShuppan (蒼樹出版), 1989. [3] H. Okumura, The arbelos in Wasan geometry, problems of Izumiya and Nait¯o,Journal of Classical Geometry, to appear. [4] H. Okumura and M. Watanabe, The twin circles of Archimedes in a skewed arbelos, Forum Geom., 4(2004), 229–251.